Multiply the following complex numbers: $({4-2i}) \cdot ({4-3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4-2i}) \cdot ({4-3i}) = $ $ ({4} \cdot {4}) + ({4} \cdot {-3}i) + ({-2}i \cdot {4}) + ({-2}i \cdot {-3}i) $ Then simplify the terms: $ (16) + (-12i) + (-8i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 16 + (-12 - 8)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 16 + (-12 - 8)i - 6 $ The result is simplified: $ (16 - 6) + (-20i) = 10-20i $